Melting Lead in a Kiln: How Long Will it Take?

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To determine how long it will take to melt 2 kg of lead in a kiln at 617.5 degrees F, the initial temperature of the lead at 55 degrees F must be converted to Celsius. The relevant equations include the heat required for melting (Q=mc) and the latent heat of fusion (Lf). The discussion highlights the need for a coefficient of heat transfer to calculate the time, as well as clarifying the constants used, such as latent heat values. Participants express confusion over the temperature units and the lack of information regarding the heat transfer rate. The consensus is that additional data is necessary to solve the problem effectively.
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Homework Statement



A girl wishes to make a lead casting of an object which will require 2kg of melted lead. if the lead was stored in her basement at approximately 55 degrees F, and it is melted to a 1/4" thick steel pot of approximate cross sectional area of 12in^2, how long will it take to melt the lead if its placed in a kiln at 617.5 degrees F, the melting point of lead?

Homework Equations



Q=mc delta t
Q=mc

Constants for lead: Lf=.25E5 J/kg Lv=8.7E5 J/kg c=130J/kgCelsius
Constants for steel: k=40 J/s times m times Celsius

The Attempt at a Solution



Q=2(130)(562.5)
Q=1.46E5
Qv=2(8.7E5)
Qv=1.74E6
Qf=2(.25E5
Qf=50000

Not sure where else to go from here...
 
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A long one to go on this I am afraid. First off your data is in Fahrenheit and your constants in celcius. What course is this--i'm thinking you have overall energy requirements posted above, but nothing about how fast the process will occur.
 
oh man...the temp units are off. a complicated problem you loose track of the basics...

I gave all of the info that i have...and its for my calc based physics class
 
Thats reassuring. To solve this will require some math. There should be some coefficient of heat transfer, etc.Not sure what Lf and all the other constants represent, the only one that I see which has a time dimension is the last.
 
lf=latent heat of fusion
lv=latent heat of vaportation?
 
Sorry, I'm not sure I can help. Someone else may be able to.

You're given a mass of lead and the shape it will become. You also know the lead is just melted and not vaporized, so know need to worry about lv. I believe the temp of the steel can be assumed to be constant, so you know exactly the number of calories it will take to make 2 Kg of solid lead to molten lead. You're given a bunch more info, but nothing else? Take an example with water--you take a liter of very cold ice and dump it into a steel pot at 0 degrees--don't you need to know something about the shape of ice block and NOT the pot, as well as rate of heat transfer. Wish I could help more, problem just doesn't have enough info imo.
 
it is alright, thank you for attempting though..
 
Any other thoughts?
 
=[...
 
  • #10
itr said:
Any other thoughts?


Wish I had.
 
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